If we assume Aa is husband which is hetero-zygote and aa is wife which is normal then their cross will be Aa X aa and first generation will be Aa aa aa Aa.
Further explanation
Cross between Aa (heterozygote male) and aa (normal female)
Here is cross between a normal wife (aa) and a heterozygote husband (Aa)
Aa X aa
First generation
Aa Aa aa aa
Result
It is clearly seen that ration is 1:1 which mean half are heterozygote and half are normal offspring. As they already told about number of offspring which is 25, So according to 1:1 12.5 are normal and 12.5 are heterozygote.
Chi-square Test
It is test used to evaluate the observed the expected values.
Chi-square formula
χ2=∑(o-e)2/e
Here o= observed and e= expected
Calculation through chi-square value
It is given below
For heterozigous (14-12.5)2/12.5=0.18
For normal (11-12.5)2/12.5=0.18
By putting values of both heterozygous and recessive in chi-square formula
χ2=0.18+0.18=0.36
Answer details
Subject: Biology
Level: High school
Key words
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The value for chi-square for this observation would be 0.36.
Further Explanation:
During gene interaction, different sets of phenotypic ratios are observed. Sometimes the observed ratio might be different from the one expected. A Chi-square test helps to compare the observed and expected value for a particular data and find whether these variables are dependent or independent of each other. The final chi-square value determines whether to accept or reject the null hypothesis.
The difference in colors of the iris in a person is called heterochromia. It usually depends on the concentration, production, and delivery of the melanin pigment in the body. It might occur as a result of genetic mosaicism, inherited from either parent, disease or injury. It is an autosomal dominant trait in which the presence of even a single abnormal dominant allele in the offspring might get him/her the disease due to the dominant character.
Here, consider the individual or father has heterochromia with heterozygous alleles (A*a) while the wife or the mother is normal-eyed and might have genotypes AA or Aa, or aa.
Where, A* = abnormal dominant allele
A = normal dominant allele
a = normal recessive allele
A cross between an individual with genotype A*a and wife with genotype AA would result in the progeny with genotypes explained in the punnett square attached below.
It is expected that 50% of the progeny would be diseased due to the presence of abnormal dominant alleles. Therefore, out of 25 offspring, 12.5 would be normal while the other 12.5 would be heterochromatic.
On the other hand, the expected values are 14 for heterochromatic offspring and 11 for normal. The table is attached below, please refer the image.
[tex]\begin{aligned}\left(\chi ^{2}\right)&=\dfrac{\sum\left(\text{O}-\text{E}\right)^{2}}{\text{E}}\\&=0.18+0.18\\&=0.36\end{aligned}[/tex]
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Answer details:
Grade: Senior School
Subject: Biology
Chapter: Biostatistics
Keywords:
Chi-square test, observed value, expected value, heterochromia, autosomal dominant disease, dominant allele, recessive allele, null hypothesis.
